No-cloning theorem does not seem preciseRigorous security proof for Wiesner's quantum moneyNo-cloning theorem and distinguishing between two non-orthogonal quantum statesProof of no-cloningHow to prove teleportation does not violate no-cloning theorem?What does it mean that copying a state is impossible but creating a copy of by entangling is possible?Cloning quantum states with a device that distinguishes between two non-orthogonal quantum statesClassical and quantum limits to classical copying?When can a fanout be used without violating the no-cloning theorem?
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No-cloning theorem does not seem precise
Rigorous security proof for Wiesner's quantum moneyNo-cloning theorem and distinguishing between two non-orthogonal quantum statesProof of no-cloningHow to prove teleportation does not violate no-cloning theorem?What does it mean that copying a state is impossible but creating a copy of by entangling is possible?Cloning quantum states with a device that distinguishes between two non-orthogonal quantum statesClassical and quantum limits to classical copying?When can a fanout be used without violating the no-cloning theorem?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
As per wikipedia, no-cloning theorem states that it is impossible to create an identical copy of an arbitrary unknown quantum state.
But from which distribution is this unknown quantum state sampled from? What does the counterfeiter know about this distribution? What is the maximum probability that a counterfeiter could succeed?
Let's say if we randomly pick one of the 2 quantum states $|psi_0big>$ and $|psi_1big>$ and give it to a counterfeiter (the states $|psi_0big>$ and $|psi_1big>$ are fixed ahead of time and are known to the counterfeiter) . The counterfeiter could measure the given state in $|psi_0big>$ and $|psi_0^perpbig>$ basis and forge with non-zero probability.
I believe success probability of a counterfeiter depends on the distribution on which the unknown state is sampled from. I would like to know for what distribution of states the no cloning theorem is applicable.
no-cloning-theorem
asked 10 hours ago
satyasatya
1413 bronze badges
$endgroup$
|
show 2 more comments
$begingroup$
As per wikipedia, no-cloning theorem states that it is impossible to create an identical copy of an arbitrary unknown quantum state.
But from which distribution is this unknown quantum state sampled from? What does the counterfeiter know about this distribution? What is the maximum probability that a counterfeiter could succeed?
Let's say if we randomly pick one of the 2 quantum states $|psi_0big>$ and $|psi_1big>$ and give it to a counterfeiter (the states $|psi_0big>$ and $|psi_1big>$ are fixed ahead of time and are known to the counterfeiter) . The counterfeiter could measure the given state in $|psi_0big>$ and $|psi_0^perpbig>$ basis and forge with non-zero probability.
I believe success probability of a counterfeiter depends on the distribution on which the unknown state is sampled from. I would like to know for what distribution of states the no cloning theorem is applicable.
no-cloning-theorem
asked 10 hours ago
satyasatya
1413 bronze badges
$endgroup$
1
$begingroup$
In your question, what are $|psi_0rangle$ and $|psi_1rangle$? Are they known to be one of $$? What do we know about $langlepsi_0|psi_1rangle$?
$endgroup$
– Mark S
9 hours ago
$begingroup$
@MarkS. I updated the question.
$endgroup$
– satya
9 hours ago
$begingroup$
If $|psi_0rangle$ were $|1rangle$ and $|psi_1rangle$ were $|+rangle$, and the counterfeiter (Eve) were given $|psi_brangle$ for a random coin $bin0,1$ unknown to her, she could not "forge with non-zero probability," correct?
$endgroup$
– Mark S
7 hours ago
$begingroup$
Or are you saying that $|psi_0rangle=alpha_0|0rangle+beta_0|1rangle$, and $|psi_1rangle=alpha_1|0rangle+beta_1|1rangle$ with $alpha_0^2$ and $alpha_1^2$ chosen uniformly at random from $[0,1]$? That seems like @AHusain's answer already addresses this.
$endgroup$
– Mark S
7 hours ago
1
$begingroup$
If the measurement output in the computational basis is $0$ then Eve knows that $b=1$. If the measurement output is $1$ in the computational basis, then she can guess, with 50% success, that $b=0$. However, she can can do better than 50%! For example, please see the answers in this question, which may be of some help.
$endgroup$
– Mark S
6 hours ago
|
show 2 more comments
$begingroup$
As per wikipedia, no-cloning theorem states that it is impossible to create an identical copy of an arbitrary unknown quantum state.
But from which distribution is this unknown quantum state sampled from? What does the counterfeiter know about this distribution? What is the maximum probability that a counterfeiter could succeed?
Let's say if we randomly pick one of the 2 quantum states $|psi_0big>$ and $|psi_1big>$ and give it to a counterfeiter (the states $|psi_0big>$ and $|psi_1big>$ are fixed ahead of time and are known to the counterfeiter) . The counterfeiter could measure the given state in $|psi_0big>$ and $|psi_0^perpbig>$ basis and forge with non-zero probability.
I believe success probability of a counterfeiter depends on the distribution on which the unknown state is sampled from. I would like to know for what distribution of states the no cloning theorem is applicable.
no-cloning-theorem
asked 10 hours ago
satyasatya
1413 bronze badges
$endgroup$
As per wikipedia, no-cloning theorem states that it is impossible to create an identical copy of an arbitrary unknown quantum state.
But from which distribution is this unknown quantum state sampled from? What does the counterfeiter know about this distribution? What is the maximum probability that a counterfeiter could succeed?
Let's say if we randomly pick one of the 2 quantum states $|psi_0big>$ and $|psi_1big>$ and give it to a counterfeiter (the states $|psi_0big>$ and $|psi_1big>$ are fixed ahead of time and are known to the counterfeiter) . The counterfeiter could measure the given state in $|psi_0big>$ and $|psi_0^perpbig>$ basis and forge with non-zero probability.
I believe success probability of a counterfeiter depends on the distribution on which the unknown state is sampled from. I would like to know for what distribution of states the no cloning theorem is applicable.
no-cloning-theorem
no-cloning-theorem
asked 10 hours ago
satyasatya
1413 bronze badges
asked 10 hours ago
satyasatya
1413 bronze badges
edited 9 hours ago
satya
asked 10 hours ago
satyasatya
1413 bronze badges
asked 10 hours ago
satyasatya
1413 bronze badges
asked 10 hours ago
satyasatya
1413 bronze badges
1413 bronze badges
1
$begingroup$
In your question, what are $|psi_0rangle$ and $|psi_1rangle$? Are they known to be one of $$? What do we know about $langlepsi_0|psi_1rangle$?
$endgroup$
– Mark S
9 hours ago
$begingroup$
@MarkS. I updated the question.
$endgroup$
– satya
9 hours ago
$begingroup$
If $|psi_0rangle$ were $|1rangle$ and $|psi_1rangle$ were $|+rangle$, and the counterfeiter (Eve) were given $|psi_brangle$ for a random coin $bin0,1$ unknown to her, she could not "forge with non-zero probability," correct?
$endgroup$
– Mark S
7 hours ago
$begingroup$
Or are you saying that $|psi_0rangle=alpha_0|0rangle+beta_0|1rangle$, and $|psi_1rangle=alpha_1|0rangle+beta_1|1rangle$ with $alpha_0^2$ and $alpha_1^2$ chosen uniformly at random from $[0,1]$? That seems like @AHusain's answer already addresses this.
$endgroup$
– Mark S
7 hours ago
1
$begingroup$
If the measurement output in the computational basis is $0$ then Eve knows that $b=1$. If the measurement output is $1$ in the computational basis, then she can guess, with 50% success, that $b=0$. However, she can can do better than 50%! For example, please see the answers in this question, which may be of some help.
$endgroup$
– Mark S
6 hours ago
|
show 2 more comments
1
$begingroup$
In your question, what are $|psi_0rangle$ and $|psi_1rangle$? Are they known to be one of $$? What do we know about $langlepsi_0|psi_1rangle$?
$endgroup$
– Mark S
9 hours ago
$begingroup$
@MarkS. I updated the question.
$endgroup$
– satya
9 hours ago
$begingroup$
If $|psi_0rangle$ were $|1rangle$ and $|psi_1rangle$ were $|+rangle$, and the counterfeiter (Eve) were given $|psi_brangle$ for a random coin $bin0,1$ unknown to her, she could not "forge with non-zero probability," correct?
$endgroup$
– Mark S
7 hours ago
$begingroup$
Or are you saying that $|psi_0rangle=alpha_0|0rangle+beta_0|1rangle$, and $|psi_1rangle=alpha_1|0rangle+beta_1|1rangle$ with $alpha_0^2$ and $alpha_1^2$ chosen uniformly at random from $[0,1]$? That seems like @AHusain's answer already addresses this.
$endgroup$
– Mark S
7 hours ago
1
$begingroup$
If the measurement output in the computational basis is $0$ then Eve knows that $b=1$. If the measurement output is $1$ in the computational basis, then she can guess, with 50% success, that $b=0$. However, she can can do better than 50%! For example, please see the answers in this question, which may be of some help.
$endgroup$
– Mark S
6 hours ago
1
1
$begingroup$
In your question, what are $|psi_0rangle$ and $|psi_1rangle$? Are they known to be one of $$? What do we know about $langlepsi_0|psi_1rangle$?
$endgroup$
– Mark S
9 hours ago
$begingroup$
In your question, what are $|psi_0rangle$ and $|psi_1rangle$? Are they known to be one of $$? What do we know about $langlepsi_0|psi_1rangle$?
$endgroup$
– Mark S
9 hours ago
$begingroup$
@MarkS. I updated the question.
$endgroup$
– satya
9 hours ago
$begingroup$
@MarkS. I updated the question.
$endgroup$
– satya
9 hours ago
$begingroup$
If $|psi_0rangle$ were $|1rangle$ and $|psi_1rangle$ were $|+rangle$, and the counterfeiter (Eve) were given $|psi_brangle$ for a random coin $bin0,1$ unknown to her, she could not "forge with non-zero probability," correct?
$endgroup$
– Mark S
7 hours ago
$begingroup$
If $|psi_0rangle$ were $|1rangle$ and $|psi_1rangle$ were $|+rangle$, and the counterfeiter (Eve) were given $|psi_brangle$ for a random coin $bin0,1$ unknown to her, she could not "forge with non-zero probability," correct?
$endgroup$
– Mark S
7 hours ago
$begingroup$
Or are you saying that $|psi_0rangle=alpha_0|0rangle+beta_0|1rangle$, and $|psi_1rangle=alpha_1|0rangle+beta_1|1rangle$ with $alpha_0^2$ and $alpha_1^2$ chosen uniformly at random from $[0,1]$? That seems like @AHusain's answer already addresses this.
$endgroup$
– Mark S
7 hours ago
$begingroup$
Or are you saying that $|psi_0rangle=alpha_0|0rangle+beta_0|1rangle$, and $|psi_1rangle=alpha_1|0rangle+beta_1|1rangle$ with $alpha_0^2$ and $alpha_1^2$ chosen uniformly at random from $[0,1]$? That seems like @AHusain's answer already addresses this.
$endgroup$
– Mark S
7 hours ago
1
1
$begingroup$
If the measurement output in the computational basis is $0$ then Eve knows that $b=1$. If the measurement output is $1$ in the computational basis, then she can guess, with 50% success, that $b=0$. However, she can can do better than 50%! For example, please see the answers in this question, which may be of some help.
$endgroup$
– Mark S
6 hours ago
$begingroup$
If the measurement output in the computational basis is $0$ then Eve knows that $b=1$. If the measurement output is $1$ in the computational basis, then she can guess, with 50% success, that $b=0$. However, she can can do better than 50%! For example, please see the answers in this question, which may be of some help.
$endgroup$
– Mark S
6 hours ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
To keep the problem small, let's say 1 qubit.
In the original statement $| psi rangle$ could be any state $alpha | 0 rangle + beta | 1 rangle$ for whatever $alpha$ and $beta$ produce a well defined state.
Quantifying over all so not a probabilistic statement. Say you uniformly draw over all possible states. It is impossible to guarantee that you have gotten $| psi rangle otimes | psi rangle$.
However if you give a probability distribution with smaller support, then it does become possible. Let's say you know the state is either $|0rangle$ or $|1rangle$ with probabilities $p$ and $1-p$ respectively. Then you could guarantee that you have gotten $| psi rangle otimes | psi rangle$. You just apply a $CNOT$ on $| psi rangle otimes | 0 rangle$.
Further example, suppose the state is either $|0rangle$, $|1rangle$, $| + rangle$ or $|-rangle$ with probabilities $p-epsilon_1$, $1-p-epsilon_2$, $epsilon_1$ and $epsilon_2$ respectively. Then you could use the same gate and it would still be likely to succeed because $epsilon_1$ and $epsilon_2$ are so small.
You could make a statement quantified like $forall | psi rangle$ in the support of a given probability distribution. Then it would depend on the distribution. If it was the first one, supported only on 2 states, then it would be possible to get $| psi rangle otimes | psi rangle$. But if it was the second it would be impossible to guarantee that result even if it would be quite likely.
answered 9 hours ago
AHusainAHusain
2,7452 gold badges4 silver badges13 bronze badges
$endgroup$
add a comment |
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$begingroup$
To keep the problem small, let's say 1 qubit.
In the original statement $| psi rangle$ could be any state $alpha | 0 rangle + beta | 1 rangle$ for whatever $alpha$ and $beta$ produce a well defined state.
Quantifying over all so not a probabilistic statement. Say you uniformly draw over all possible states. It is impossible to guarantee that you have gotten $| psi rangle otimes | psi rangle$.
However if you give a probability distribution with smaller support, then it does become possible. Let's say you know the state is either $|0rangle$ or $|1rangle$ with probabilities $p$ and $1-p$ respectively. Then you could guarantee that you have gotten $| psi rangle otimes | psi rangle$. You just apply a $CNOT$ on $| psi rangle otimes | 0 rangle$.
Further example, suppose the state is either $|0rangle$, $|1rangle$, $| + rangle$ or $|-rangle$ with probabilities $p-epsilon_1$, $1-p-epsilon_2$, $epsilon_1$ and $epsilon_2$ respectively. Then you could use the same gate and it would still be likely to succeed because $epsilon_1$ and $epsilon_2$ are so small.
You could make a statement quantified like $forall | psi rangle$ in the support of a given probability distribution. Then it would depend on the distribution. If it was the first one, supported only on 2 states, then it would be possible to get $| psi rangle otimes | psi rangle$. But if it was the second it would be impossible to guarantee that result even if it would be quite likely.
answered 9 hours ago
AHusainAHusain
2,7452 gold badges4 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
To keep the problem small, let's say 1 qubit.
In the original statement $| psi rangle$ could be any state $alpha | 0 rangle + beta | 1 rangle$ for whatever $alpha$ and $beta$ produce a well defined state.
Quantifying over all so not a probabilistic statement. Say you uniformly draw over all possible states. It is impossible to guarantee that you have gotten $| psi rangle otimes | psi rangle$.
However if you give a probability distribution with smaller support, then it does become possible. Let's say you know the state is either $|0rangle$ or $|1rangle$ with probabilities $p$ and $1-p$ respectively. Then you could guarantee that you have gotten $| psi rangle otimes | psi rangle$. You just apply a $CNOT$ on $| psi rangle otimes | 0 rangle$.
Further example, suppose the state is either $|0rangle$, $|1rangle$, $| + rangle$ or $|-rangle$ with probabilities $p-epsilon_1$, $1-p-epsilon_2$, $epsilon_1$ and $epsilon_2$ respectively. Then you could use the same gate and it would still be likely to succeed because $epsilon_1$ and $epsilon_2$ are so small.
You could make a statement quantified like $forall | psi rangle$ in the support of a given probability distribution. Then it would depend on the distribution. If it was the first one, supported only on 2 states, then it would be possible to get $| psi rangle otimes | psi rangle$. But if it was the second it would be impossible to guarantee that result even if it would be quite likely.
answered 9 hours ago
AHusainAHusain
2,7452 gold badges4 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
To keep the problem small, let's say 1 qubit.
In the original statement $| psi rangle$ could be any state $alpha | 0 rangle + beta | 1 rangle$ for whatever $alpha$ and $beta$ produce a well defined state.
Quantifying over all so not a probabilistic statement. Say you uniformly draw over all possible states. It is impossible to guarantee that you have gotten $| psi rangle otimes | psi rangle$.
However if you give a probability distribution with smaller support, then it does become possible. Let's say you know the state is either $|0rangle$ or $|1rangle$ with probabilities $p$ and $1-p$ respectively. Then you could guarantee that you have gotten $| psi rangle otimes | psi rangle$. You just apply a $CNOT$ on $| psi rangle otimes | 0 rangle$.
Further example, suppose the state is either $|0rangle$, $|1rangle$, $| + rangle$ or $|-rangle$ with probabilities $p-epsilon_1$, $1-p-epsilon_2$, $epsilon_1$ and $epsilon_2$ respectively. Then you could use the same gate and it would still be likely to succeed because $epsilon_1$ and $epsilon_2$ are so small.
You could make a statement quantified like $forall | psi rangle$ in the support of a given probability distribution. Then it would depend on the distribution. If it was the first one, supported only on 2 states, then it would be possible to get $| psi rangle otimes | psi rangle$. But if it was the second it would be impossible to guarantee that result even if it would be quite likely.
answered 9 hours ago
AHusainAHusain
2,7452 gold badges4 silver badges13 bronze badges
$endgroup$
To keep the problem small, let's say 1 qubit.
In the original statement $| psi rangle$ could be any state $alpha | 0 rangle + beta | 1 rangle$ for whatever $alpha$ and $beta$ produce a well defined state.
Quantifying over all so not a probabilistic statement. Say you uniformly draw over all possible states. It is impossible to guarantee that you have gotten $| psi rangle otimes | psi rangle$.
However if you give a probability distribution with smaller support, then it does become possible. Let's say you know the state is either $|0rangle$ or $|1rangle$ with probabilities $p$ and $1-p$ respectively. Then you could guarantee that you have gotten $| psi rangle otimes | psi rangle$. You just apply a $CNOT$ on $| psi rangle otimes | 0 rangle$.
Further example, suppose the state is either $|0rangle$, $|1rangle$, $| + rangle$ or $|-rangle$ with probabilities $p-epsilon_1$, $1-p-epsilon_2$, $epsilon_1$ and $epsilon_2$ respectively. Then you could use the same gate and it would still be likely to succeed because $epsilon_1$ and $epsilon_2$ are so small.
You could make a statement quantified like $forall | psi rangle$ in the support of a given probability distribution. Then it would depend on the distribution. If it was the first one, supported only on 2 states, then it would be possible to get $| psi rangle otimes | psi rangle$. But if it was the second it would be impossible to guarantee that result even if it would be quite likely.
answered 9 hours ago
AHusainAHusain
2,7452 gold badges4 silver badges13 bronze badges
answered 9 hours ago
AHusainAHusain
2,7452 gold badges4 silver badges13 bronze badges
answered 9 hours ago
AHusainAHusain
2,7452 gold badges4 silver badges13 bronze badges
answered 9 hours ago
AHusainAHusain
2,7452 gold badges4 silver badges13 bronze badges
2,7452 gold badges4 silver badges13 bronze badges
add a comment |
add a comment |
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$begingroup$
In your question, what are $|psi_0rangle$ and $|psi_1rangle$? Are they known to be one of $$? What do we know about $langlepsi_0|psi_1rangle$?
$endgroup$
– Mark S
9 hours ago
$begingroup$
@MarkS. I updated the question.
$endgroup$
– satya
9 hours ago
$begingroup$
If $|psi_0rangle$ were $|1rangle$ and $|psi_1rangle$ were $|+rangle$, and the counterfeiter (Eve) were given $|psi_brangle$ for a random coin $bin0,1$ unknown to her, she could not "forge with non-zero probability," correct?
$endgroup$
– Mark S
7 hours ago
$begingroup$
Or are you saying that $|psi_0rangle=alpha_0|0rangle+beta_0|1rangle$, and $|psi_1rangle=alpha_1|0rangle+beta_1|1rangle$ with $alpha_0^2$ and $alpha_1^2$ chosen uniformly at random from $[0,1]$? That seems like @AHusain's answer already addresses this.
$endgroup$
– Mark S
7 hours ago
1
$begingroup$
If the measurement output in the computational basis is $0$ then Eve knows that $b=1$. If the measurement output is $1$ in the computational basis, then she can guess, with 50% success, that $b=0$. However, she can can do better than 50%! For example, please see the answers in this question, which may be of some help.
$endgroup$
– Mark S
6 hours ago